Chapter 2- Constant Speed and Accelerated Motion

Class Notes 9/6/11:

Basic Terms:

• Position
  • The exact location of something (using reference points)
• Distance
  • The total amount (length) traveled from start to finish; how far
• Displacement
  • The progress made traveling; must include a direction; change in position
• Speed
  • How fast you're going
• Velocity
  • How fast you're going + direction
• Vector
  • Quantity that has direction and size
• Scalar
  • Quantity that is size only
• Rate of change of position
  • v=d/t
• Motion Diagrams:

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Lesson 1- Kinematics: describing motion with words
Hw Wed 9/7/11
Lesson 1 Physics Classroom (a, b, c, d)
Physics Classroom

1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
   1. There were several topics within the reading that I was familiar with due to our class discussion. The first example of this is the difference between scalar and vector quantities. I know that scalar quantities solely focus on a number (size), whereas vector quantities include both the numerical value and a direction. Speed and velocity were also discussed in class, and I understand that speed is a scalar quantity, whereas velocity is a vector quantity.Additionally, I was able to already understand the difference between distance and displacement, for this too was discussed during class. Distance is simply the total length traveled from start to finish, a scalar quantity, and displacement is just the progress made and change in position, which must include a direction, and thus a vector quantity.
2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
   1. Initially, I was a little confused about the relationship between speed and velocity. The reading helped me understand that an object may be moving at 25 m/s, which would be it's speed, but if one were to say it was moving at 25 m/s east, this would be it's velocity. Therefore, the two are the same, except for the fact that velocity includes the direction. The website simply and clearly states, "velocity is speed with direction", which furthered my understanding of the concept.
3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
   1. If an object's velocity is decreasing, would this be considered deceleration, or would it still be acceleration, but possibly have a negative (downward) value?
4. What (specifically) did you read that was not gone over during class today?
   1. Acceleration was the main topic that was in the reading, which was not discussed in class thus far. The reading explains that acceleration is the rate at which the velocity of an object changes, and is therefore a vector quantity. The "rule of thumb" for calculating the direction of of the acceleration vector is if the object is slowing down, then it's acceleration will be in the opposite direction (+ or -) of it's motion. Due to this, there can be both positive and negative accelerations. Since the average acceleration can be calculated by diving the change in velocity by the time, the units are the units of velocity, plus time, such as m/s/s, or m/s^2.

Notes 9/8/11:
At rest= not moving at all (v=0)
Constant Speed= the speed is non-changing (ex: always 50)
Average Speed= the average of all of the speeds (ex: sometimes 48, 52, so average is 50)
Instantaneous Speed= the speed at a given instance/time
(for all, v=d/t)

Increasing speed= speeding up
Decreasing speed= slowing down

Motion Diagram (arrows)
Dot/ ticker tape Diagram
- makes sparks

Lesson 2: Describing Motion with Diagrams
HW Thurs 9/8
Lesson 2 Physics Classroom (a, b, c)

1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
   1. From the class discussion today, I learned the basics about ticker tapes, including how and why they are used. Common in lab procedures, a ticker tape analysis, also known as an oil drop diagram, enables one to see the objects motion over time. For example, if an object is moving at a constant speed, the dots will be the same distance apart, yet if it's speed increases, these dots will become increasingly farther apart. During class, vector diagrams were also briefly discussed. These are pictures of arrows that depict the velocity of an object as it is moving, using arrows. These arrows can change size depending on the speed. In class I also learned that for increasing speed, the arrow for acceleration points in the same direction as the velocity arrows, yet for decreasing speed vector diagrams, the acceleration arrow points in the opposite direction as the velocity arrow.
2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
   1. During class, I was a little uncertain as to the connection between using a ticker tape and the acceleration of the object. I didn't fully understand how analyzing such a diagram would enable you to determine whether or not the object was moving with a constant velocity, or accelerating. After reading over the physics classroom page, I now understand that the distance in between the dots on the ticker tape diagram include this information. If the dots are an equal distance apart, then it has a constant velocity, whereas if the distance between the dots changes, either more distance or less, this means that the velocity is changing, and thus the object is accelerating.
3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
   1. there are no questions I have pertaining to these sections of the reading.
4. What (specifically) did you read that was not gone over during class today?
   1. On the website, I read that vector diagrams can be used to analyze not only velocity and acceleration, but also force, momentum, and other physical quantities. I also learned that a large distance between dots indicated that it is moving quickly, and if the dots are very close together, that means the object is moving slowly.

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LAB: Speed of a CMV (constant motion vehicle)

Lauren Kostman
Lab Partner: Jonathan Itskovitch
September 9, 2011

Objectives (3):
1.) How precisely can you measure distances with a meter stick?
2.) How fast does your CMV move?
3.) What information can you get from a position-time graph?
Materials:
- spark timer and spark tape
- meter stick
- masking tape
- CMV
Hypotheses:
1.) You can measure distances with a meter stick up to the millimeter.
2.) My CMV moves approximately 5 miles an hour.
3.) From a position-time graph, you can get the instantaneous speeds for the CMV, and then calculate the average speed. Furthermore, you can also get the velocity due to the direction of the position (of the CMV) depicted on the graph.
Length of Laptop: 32.90
Data Table- Position versus Time:

Position-time Graph:

Analysis:
The equation of the position-time graph shown above is y=59.985x. In this equation, the slope is represented by the 59.985. In other words, this is the change in position (cm) divided by the change in time (sec). Since the speed is constant, and the slope represents the speed, the slope is constant, too. Furthermore, the graph shows an increasing line from left to right,which proves that the speed of the CMV was moving in a positive direction (away), at a constant speed.
Discussion Questions:
1.) Why is the slope of the position-time graph equivalent to average velocity?
The slope of the position-time graph is equivalent to average velocity because slope is delta y divided by delta x, and in the position-time graph, we are dividing the change in position (which is on the y-axis) by the change in time (which is on the x-axis). Therefore, the slope is equivalent to the average velocity, for slope =cm/sec(speed).
2.) Why is it average velocity and not instantaneous velocity? What assumptions are we making?
It is average velocity as opposed to instantaneous velocity because rather than focusing on the velocity at a given moment, or instance, we averaged the velocities instead. We are assuming that the car travels at a constant speed, since there is a constant slope, and that our calculations have averaged an accurate average velocity of the CMV.
3.) Why was it okay to set the y-intercept equal to zero?
It was okay to set the y-intercept to zero because this represents position, and it would be impossible for the CMV to begin at a negative position. Due to this, the y-intercept was placed at 0, the starting position of the vehicle.
4.) What is the meaning of the R^2 value?
The meaning of the R^2 value is how close the data is to actually making a straight line, and abiding by the linear equation, with the constant slope. In the position-time graph, R^2 equals .99461, which is extremely close to a perfectly straight line (every point is on the line).
5.) If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours?
If I were to add the graph of another CMV that moved more slowly on the same axes as my current graph, I would expect it to have a smaller slope, for this is equivalent to speed as aforementioned, and thus be less steep. In other words, it would lie below the faster car's line.
Conclusion:
Subsequent to completing the lab pertaining to the speed of a constant moving vehicle, my lab partner and I were able to determine the speed of our vehicle, which is approximately 59.99 cm/s. Due to this, our second hypothesis- that the vehicle travels at about five cm/s, is tremendously incorrect. However, it was hypothesized that a meterstick can be utilized to measure distances precisely to the centimeter, which is also inaccurate, for the meter stick used in this lab measured only up to the second decimal place, the centimeter. Upon completing the position-time graph, we learned that from this, we can see the instantaneous velocities, and then calculate the average velocity. This supports our third and final hypothesis, since we were able to get a graphed version of our data from analyzing such a graph. There are several sources of error that may have contributed to inaccuracies, including the fact that the beginning speeds were ignored, and thus we may have either misjudged too many or too few of the dots, which could harm the results. Additionally, it's possible that the point at the zero cm mark shifted, which would also effect the data results. If one were to redo this lab, these issues could be minimized by using the last ten dots, this way there wouldn't be the chance of ripping off the wrong amount of dots at the beginning, and also by either taping the ticker tape to the meter stick so a shift could not occur, or by simply using measuring tape, so that it's a flatter surface and there's less of a chance for a shift to occur.

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Graphing:

Lauren Kostman
Ms. Burns
Honors Physics
September 12, 2011

All Graphs:


Graph: Walking Towards

Graph: Walking Away

Graph: At Rest (from about 6 sec to 8 sec)

Extra Graph (caught arm movement):


Discussion Questions:

1. How can you tell that there is no motion on a…
   1. Position vs. time graph
      1. There would be no slope; it would just be a horizontal line across the y-axis.
   2. Velocity vs. time graph
      1. There would be no slope, it would be a horizontal line. However, the velocity would equal zero, so the line would be at the x-axis.
         
         
   3. Acceleration vs. time graph
      1. There would be a horizontal line at zero.

1. How can you tell that your motion is steady on a…
   1. Position vs. time graph
      1. There would be a constant slope- it would remain the same throughout.
   2. Velocity vs. time graph
      1. The velocity would remain the same (constant) on a straight line, but ‘y’ would not be equal to zero.
         
         
   3. Acceleration vs. time graph
      1. The slope would be zero (straight line at 0).

1. How can you tell that your motion is fast vs. slow on a…
   1. Position vs. time graph
      1. It would be faster if the slope is steeper and slower if it is less steep.
   2. Velocity vs. time graph
      1. It would be faster if it is more positive/negative, and slower if it is less positive/ negative.
         
         
   3. Acceleration vs. time graph
      1. You wouldn't know- you can go fast with a slower acceleration or vice versa, therefore the acceleration versus time graph would not indicate if the motion is fast or slow.

1. How can you tell that you changed direction on a…
   1. Position vs. time graph
      1. If you are walking away from the sensor, it is a positive slope, but if you are walking away from it, it is a negative slope.
   2. Velocity vs. time graph
      1. If you are walking straight (in a positive direction) the slope of the would be positive- increasing from left to right, and if you are walking backwards (in a negative direction), the slope of the line would be negative, and thus decrease from left to right.
         
         
   3. Acceleration vs. time graph
      1. On such a graph, a change in direction would be hard to tell for this graph goes up and down even if the direction doesn't change, thus determining this would be nearly impossible.

1. What are the advantages of representing motion using a…
   1. Position vs. time graph
      1. This is a simple way to analyze how far you have gone in a certain amount of time.
   2. Velocity vs. time graph
      1. This type of graph allows you to see the change in both speed and direction.
   3. Acceleration vs. time graph
      1. This graph shows if someone's speed in increasing or decreasing, and also shows if there's a change in velocity, or if the velocity remains constant.

1. What are the disadvantages of representing motion using a…
   1. Position vs. time graph
      1. The data on this particular graph may not always be accurate, for if a person is swaying his or her arms, this movement may be caught by the sensor and cause an error in the data, and then the results will be off.
   2. Velocity vs. time graph
      1. The information may not always be correct if an error occurs, such as if the sensor is pointing too low at the person's body, it will catch the movement of the legs, not the torso, and then give inaccurate results.
   3. Acceleration vs. time graph
      1. From this graph, the least amount of information can be determined, for it just shows the rate of change in velocity, but doesn't include the posittion or the velocity.

1. Define the following:
   1. No motion
      1. No motion is when the object is at rest; no distance or displacement occurs here.
   2. Constant speed
      1. Constant speed is when an object is moving (the same amount of distance) at the same rate of time. Thus, at each given instant, the same distance has been covered over the same period of time.

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September 13, 2011
Graphs: at rest versus constant speed

Graphs: increasing versus decreasing speed

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Classnotes:

"The Big 5"

(Top 5 most common equations)

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Classwork- Wednesday September 14:
Graph: moving away; sensor at bottom

Graph: moving away

Graph: moving towards

Graph: moving toward; sensor at bottom

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Lauren Kostman
HW 9/14/11
Physics Classroom

Lesson 3: Describing Motion with Positive vs. time Graphs

1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
   1. One concept that was discussed in class today and present in the reading, as well, was that in a position versus time graph, the line will be straight if it's increasing (or decreasing) at a constant rate (slope), but will appear curved if the slope is not constant, and is thus changing. This can be exemplified by the following graphs:

Furthermore, the fact that slope represents velocity (and speed) was also discussed in class. From viewing the graph, one is able to calculate the slope by finding two points on the graph and plugging those points into the slope equation, which is the change in y divided by the change in x, known as the "rise over run" formula. By having the slope, one knows that this is equivalent to the velocity; a positive slope equals a positive velocity and vice versa.
2.) What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.

1. The reading helped me understand that the slower an object moves, the flatter it's line will appear, whereas a faster object will have a much steeper slope. Initially, this was a little unclear to me, but the reading included examples that enabled me to understand the difference between the two. In other words, if an object is moving very slowly, it's velocity will be lower, and thus it will have a smaller slope, causing it to be less steep.The following graphs exemplify this:

3.) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
- I don't have any specific questions so far.
4.) What (specifically) did you read that was not gone over during class today?
- A topic that I read online that wasn't specifically discussed in class today was positive/negative acceleration. Negative acceleration is when an object moves in the negative direction and speeds up, whereas positive acceleration is when an objects moves in the negative direction and slows down. The graph on the left represents negative acceleration, and the graph on the right represents positive acceleration:

Lesson 4: Describing Motion with Velocity vs. Time Graphs

1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
   1. One thing that was covered in this section that I understood from class was that if an object is moving with a positive velocity and zero acceleration, the graph would appear like a horizontal line above zero (in the positive region of the graph). Furthermore, if an object moves with a positive velocity and has a positive acceleration, the graph will appear like a straight line, increasing from left to right in the positive region of the graph. Additionally, I already understood that on a velocity versus time graph, the acceleration equals the slope. For example, if an object moves with a constant velocity, it won't have any acceleration, and thus it's slope will equal zero (it will be a horizontal line).
2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
   1. The reading helped me understand the concept of how to fully interpret and analyze such graphs. I was a bit confused as to what factors represented what appearances of the graph. For example, I now understand that a positive (rightward) velocity means that positive velocity values are plotted, and a rightward, or positive, acceleration means that the slope is positive. Yet, if the object has a rightward velocity but a leftward acceleration, this means it decreases from left to right (negative slope, negative acceleration). If an object has a leftward (-) velocity, this would mean that it is in a negative region of the graph.
3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
   1. I don't have any specific questions thus far.
4. What (specifically) did you read that was not gone over during class today?
   1. In class, it was briefly mentioned that the area between the line and the x-axis represents displacement. However, it wasn't fully covered as to how one would calculate this. In the reading, I learned that there are three methods of approach, each relating to a different linear situation (ex: horizontal line, line increasing, etc). If the area shaded is a rectangle, you find the area using the A=bh formula; if the area if triangular, one would use the A=.5bh formula, and lastly, if it is a trapezoid shape, the area formula to find the displacement would be .5b(h1+h2).

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Position-Time Graphs
September 15, 2011
A.)

1.) Qualitative Description (final-initial position):
A-B: at rest
B-C: constant, towards
C-D: at rest
D-E: constant, negative direction (behind origin)
E-F: at rest
F-G: constant, towards origin (+ direction)
G-H: constant, away from origin (+ direction)
2.) Displacement:
A-B: 0
B-C: -10
C-D: 0
D-E: -16
E-F: 0
F-G: +16
G-H: 14
3.) Velocity (m/s):
A-B: 0
B-C: -10/2= -5
C-D: 0
D-E: -16/.5= -32
E-F: 0
F-G: 16/1= 16
G-H: 14/2= 7
4.) Average Speed:
v=d/t
v= total distance/total time
d= add up displacements; d= 56
t= 11 sec
v= 56/11
v= 5.1 m/s
5.) Average Velocity:
v= change in position/ total time
d=14-10=4
t= 11
v= 4/11
v= .44 m/s
6.) Acceleration (for each segment):
A-B: 0
B-C: 0
C-D: 0
D-E: 0
E-F: 0
F-G: 0
G-H: 0
7.) Draw as a velocity-time graph

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LAB: Acceleration Graphs

Lauren Kostman
lab partner- Jonathan Itskovitch
September 16, 2011

Objectives:

• What does a position-time graph for increasing speeds look like?
• What information can be found from the graph?

Available Materials:
Spark tape, spark timer, track, dynamics cart, ruler/meterstick/measuring tape
Important Notes:

• Be sure to write your hypothesis BEFORE doing the experiment.
• All data should be recorded in an organized spreadsheet in Excel.
• You need to document your procedure. Decide with your lab partner how you intend to do this.

Hypotheses:
1.) A position-time graph for increasing speed would become steeper, towards the right; the slope wouldn't be constant.
2.) From the graph, the velocity from different times and the average velocity can be determined.
Procedure:
1.) Acquire spark tape, spark timer, track, dynamics cart, ruler/meterstick/measuring tape
2.) Set up experiment by plugging in the spark timer to an outlet
3.) Place a ramp on top of a textbook, with the end of the book lining up with the end of the ramp

4.) Tape the ticker tape to the cart (using masking tape)

5.) Insert the ticker tape into the spark timer (make sure it is set to 10 hertz)
6.) There will be 2 directions- away and towards- that the cart will be pushed in (when the car moves in the positive direction, it starts at the top with the ticker tape pulled through, and when it moves in the negative direction, it starts at the bottom with the ticker tape pulled through)

7.) Push the car in the 2 indicated directions above


8.) Measure the distance between the dots on all four strips of ticker tape using a measuring tape

9.) Place this information into a data chart
10.) Interpret this data to make a position-time graph (using an Excel spreadsheet)

Analysis:
a) Interpret the equation of the line (slope, y-intercept) and the R2 value.
The equation of the line tells us the position, initial velocity, average velocity, and average acceleration. The y-value indicates the (change in) position of the graph (at a given time). The a-value tells us half of the average acceleration; by multiplying this value by two, one would calculate the total average acceleration. The b-value represents the initial velocity. The slope of the line is the velocity at that given point. The r^2 value tells us how accurate the parabolic line is (as a percent). The x^2 value is the change in time.
b) Find the instantaneous speed at halfway point and at the end.

Increasing speed:
Instantaneous speed halfway point: 40 cm/s
Instantaneous speed at the end:-5 cm.s
Decreasing speed:
instantaneous speed halfway point: 25 cm/s
instantaneous speed at the end: 80 cm/s
c) Find the average speed for the entire trip.
The average speed for the decreasing speed graph is 31.82 cm/s.
The average speed for the increasing speed graph is 30.27 cm/s.

Discussion Questions:

1. What would your graph look like if the incline had been steeper?
   1. If the incline had been steeper, the line of the graph would appear to be more vertical, heading in the upward direction. The decreasing speed graph would have a more vertical line, too, but heading in the downward direction.
2. What would your graph look like if the cart had been decreasing up the incline?
   1. If the cart had been decreasing up the incline, the graph would have an initial acceleration that was high, and then it would lessen in a negative direction, eventually reaching and passing zero. When this occurs, the cart would go back down, in a faster, negative direction. Due to this, the graph would look like an upside down (negative) parabola.
3. Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
   1. On the increasing speed graph, the midpoint has a velocity of +25 cm/s. The average velocity is +30.27 cm/s. Thus, the average velocity is a bit faster than the velocity of the midpoint.
   2. On the decreasing speed graph, the midpoint has a velocity of 40 cm/s. The average velocity is 31.82 cm/s. Therefore, the average velocity is slower than that of the midpoint.
4. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?
   1. The instantaneous speed is the slope of the tangent line because it is impossible to find the change, or delta, of a point (there is none there). The tangent line represents the slope of just one point, and it doesn't come in contact with any other point on the graph. So, this is the speed of just a single point, thus making it instantaneous.
5. Draw a v-t graph of the motion of the cart. Be as quantitative as possible.

Conclusion:
Upon completing this lab, one may learn that a position-time graph for increasing speeds (moving away, in positive direction) looks like like right (positive) half of a parabola, with an increasing slope as it progresses (no constant slope).For decreasing speeds, moving in the negative direction (towards), the graph appears like the other, negative half of the parabola, with a negative, non-constant slope. Due to this, the first hypothesis regarding the looks of an increasing speed position-time graph was correct. Upon analyzing such a graph, one is able to see (or calculate) the change in position, the average acceleration, the change in time, and the initial velocity.For example, for one of the graphs, the equation was y=16.044x^2+1.3769x; the 16.044 repres Hence, the second hypothesis held some truth, but did not include all of the information one may get from such a graph. A possible source of error could have been the way the distance between dots on the ticker tape was measured; although a meterstick was used, there are other, more accurate and precise ways to measure, such as by using a measuring tape. Occasionally, metersticks are slightly warped, and as a result, the measurements are a little off; by using a measuring tape, one could line the ticker tape up against it, thus allowing the measurements to be much more accurate and precise. Additionally, the points chosen for the ticker tape, especially for the decreasing speed one, may have had inaccuracies, for the points may have been chosen on the tape before the speed was actually decreasing, and thus the results would not be correct. For future times, one can avoid this mistake by using the last 10 dots on the end.

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Interpreting Position-time Graphs (packet)
Homework
September 20, 2011

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Lauren Kostman
September 21, 2011
Catching Up/ Multi-Segment Practice Problems

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LAB: Crash Course

Lauren Kostman
Partners: Jonathan Itskovich, Gabby Leibowitz, Rob Kwark
September 23, 2011

Materials:
Constant Motion Vehicle, Tape measure and/or metersticks, Masking tape (about 30 cm/group), Stop watch, spark timer and spark tape
Hypotheses (based on calculations):
1.) It is hypothesized that in situation A, the two cars with collide in 6.54 seconds, at 204.84 cm.
2.) It is hypothesized that in situation B, the two cars will crash in 3.46 seconds, at 207.55 cm.


Vblue (cart 2)= -59.9 cm/s; y=-59.9x + 600
Vyellow (cart 1)=31.1 cm/s; y=31.1x
From the graphs (as shown above), the following equations can be made by setting the equations of the lines equal to each other:
Crashing:
-59.9x+600=31.1x
x=6.59 sec
y= 204.8 cm

Catching Up:
59.9x=31.1x+100
x=3.46 sec
y=207.5 cm

Procedure:


Situation A:

Situation B:

Results:
Situation A (cars moving towards one another)

Trial	Position of Collision(cm)
1	200 cm
2	204 cm
3	208 cm
4	210 cm
5	205 cm

Situation B (cars in same direction)

Trial	Position of Collision (cm)
1	198 cm
2	210 cm
3	205 cm
4	211 cm

Discussion questions

1. Where would the cars meet if their speeds were exactly equal?
   1. If the speeds of the two cars were exactly equal, they would meet at the midway point of the total distance traveled. So, if they were moving at the same speed in Situation A, they would meet at 300 cm, or 3 meters. In Situation B, if the speeds were the same, they never would crash because one car would have a farther start, and thus be ahead of the other car, preventing it from ever catching up at the same speed.
2. Sketch position-time graphs to represent the catching up and crashing situations. Show the point where they are at the same place at the same time.

Catching up:

y=59.9x
y=31.1x+100
Intersect (same place at same time): (3.46, 207.5)
Crashing:

y= -59.9x+600
y=31.1 x
Intersect (same place at same time): (6.59, 204.8)
3. Sketch velocity-time graphs to represent the catching up situation. Is there any way to find the points when they are at the same place at the same time?
Catching Up:

y= 59.9x
y=31.1x+100
Crashing:

y= -59.9x+600
y=31.1x
From just viewing these velocity-time graphs, there is no way to tell where the cars crash or meet up (when they're at the same place at the same time). This is because velocity-time graphs just show the velocity of the object at a given amount of time. For both Situation A and B, each car was moving at a constant speed, and that is why the velocity-time graphs have straight, horizontal lines. There is no acceleration, and thus no slope.
Analysis:
Overall, the results from this experiment were very good. Good results means that the percent error is less than ten percent. Precise results are when the results are all of very close, or equal values. This can be determined by using the percent difference formula.

For Situation A:

Trial	Percent Error	Percent Difference
1	2.4 %	2.6 %
2	.4 %	.7%
3	1.5 %	1.3 %
4	2.5 %	2.2 %
5	0.1 %	0.2 %

Calculations (trial 1)
% error
theoretical= 204.84 cm
experimental= 200 cm
% error= (204.84-200)/204.84 x 100
= 4.84/204.84x 100
= 2.4 %
(same calculations for each, but fill in individual trial result- not 200 each time-)
% difference
average experimental value= 205.4 cm (sum of all trial values divided by 5)
individual experimental value= 200 cm
% difference= (205.4-200)/205.4 x 100
= 2.6 %
(same calculations for each, but substitute 200 for individual trial result)

For Situation B:

Trial	Percent Error	Percent Difference
1	4.6 %	3.9 %
2	1.2 %	1.9 %
3	1.2 %	0.5 %
4	1.7 %	2.4 %

Calculations (trial 1)
% error:
theoretical= 207.55 cm
experimental= 198 cm
% error= (207.55-198)/207.55 x 100
= 4.6 %
(same calculations for each trial, but replace the 198 for each separate experimental value)
% difference
average experimental value= 206 cm (add up experimental values and divide by 4)
individual experimental value= 198 cm
% difference= (206-198)/206 x 100
= 3.9 %
(same calculations for each but replace the 198 for each individual experimental value)

These calculations prove that for both Situations, A and B, the results were very precise, and accurate as well.
Conclusion:
In this lab, no formal hypothesis were made; however, before conducting the experiment, it was calculated that in Situation A, the two cars would crash at 207.55 cm, at 3.46 seconds. The results from the lab support this calculation, for all of the trials came within less than a 2.5% error. In Situation B, it was calculated that the cars would meet up at 204.84 cm, at 6.59 seconds. Upon completing all of the trials, the results in this case, as well, seem to support this calculation, for all of the results came under a 5% error. Nonetheless, there are some errors that could have impacted the results from this lab, including the fact that the speeds of each of the cars were tested a few weeks ago, and thus the batteries may have been stronger at that point, enabling the cars to move faster than they did the day of the crash course lab. A way to prevent this error in the future is to either insert new batteries into the cars before performing the experiment, or by simply recalculating the average speeds of the cars before conducting the experiment. Another error could have been the lack of precision while measuring the crashing points; although the point of collision was recorded, it's possible that the exact location of the crash was not the same as where the person recording it believed. A way to prevent this error form occurring is to perhaps record (using a video camera) the crash from close up, this way the exact location of the crash can be lined up with the exact measurement (on the ramp), enabling the results to be more accurate.

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Egg Drop Project

Lauren Kostman
Partner: Jonathan Itskovich
September 28, 2011

Pictures of Final Project:


Description of Final Project:
The methods used for creating the final design for the project included creating a light weight yet sturdy apparatus. Initially, we created a paper cube with 4 springs inside, all of which were connected to a paper cylinder that held the egg inside. For several reasons, this prototype failed. The metal springs were very heavy, causing a large (faster) acceleration, and there was not enough support for the egg, thus when dropped, it became "scrambled". After realizing the issues with the first prototype, a completely new one was made, which consisted of a woven stacked straw base, a tin-foil "cup"/ mold for the egg to sit in, straws surrounding the tin foil, a rubber band holding the tops of the straws together, and a paper parachute connected by sewing thread on the top. This new mechanism was very lightweight, and the parachute enabled the acceleration to lessen (slowed it down). Therefore, dropping it was a success, for the egg remained whole.

Results:
The egg remained whole (in tact) after the fall.

Calculation of Acceleration:
d= 8.5 m
vi= 0 m/s
t= 2.4 s
a=?

d= vit + .5at^2
8.5=0(2.4)+.5(2.4)^2
8.5= 2.88a
a=2.95 m/s/s

Comparison of Acceleration to 9.8 m/s/s:
Without any help, the acceleration of the egg (due to gravity) would be about 9.88 m/s/s. However, using our design, the acceleration was able to decrease to 2.95 m/s/s, and thus it didn't crack. A slower acceleration, in this case, is helpful for it lessens the hard impact upon landing.

Future Changes:
If I were to redo this project, something I would do differently is try using less tin foil and more straws, for this would make the apparatus weigh even less, and thus lessen the acceleration that much more. Although tin foil is pretty light weight, it's still a bit heavier than straws, and many of the other models that were made of just straws seemed to be successful. Nonetheless, the final mechanism created was very successful, so upon completing this project again, I would definitely use a parachute and straws again.

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Free-Fall

Lauren Kostman
October 3, 2011
Class Notes

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Lauren Kostman
October 3, 2011
Homework- Physics Classroom: Lesson 5
Free Fall and the Acceleration of Gravity

Introduction to Free Fall
Topic Sentence: During free-fall, all objects move at the same acceleration (-9.8m/s/s), and do not encounter air resistance.
A free falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects:

• Free-falling objects do not encounter air resistance.
• All free-falling objects accelerate downwards at a rate of 9.8 m/s/s

Ticker tape or dot diagram would depict acceleration during free-fall.
The Acceleration of Gravity
Topic Sentence: During free-fall, objects accelerate downwards by a value of g, which is the acceleration of gravity.
A free-falling object has an acceleration of 9.8 m/s/s, downward; this value is known as the acceleration of gravity (g).
The value of g is different in different gravitational environments.

Acceleration is the rate at which an object changes its velocity. To accelerate at 9.8 m/s/s means to change the velocity by 9.8 m/s each second.

The object’s velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s/s.
Another way to represent this acceleration of 9.8 m/s/s is to add numbers to the dot diagram.
Representing Free Fall by Graphs
Topic Sentence: Graphs can be used to analyze the motion of an object during free-fall.
One way of describing the motion of objects is through the use of graphs.

The line on the graph curves, which signifies an accelerated motion (-9.8m/s/s). The object starts with a small velocity (slow) and finishes with a large velocity (fast). The negative slope of the line indicates a negative (i.e., downward) velocity.

The line on the graph is a straight, diagonal line, which signifies an accelerated motion (g = 9,8 m/s/s, downward). The object starts with a zero velocity and finishes with a large, negative velocity (it’s moving in the negative direction and speeding up (so it has a negative acceleration). The constant, negative slope indicates a constant, negative acceleration.

How Fast? and How Far?
Topic Sentence: Formulas can be used to calculate the velocity and distance of an object during free-fall.
Free-falling objects are in a state of acceleration (9.8 m/s/s). Each second, add 9.8 m/s to velocity. The velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen. The formula for determining the velocity of a falling object after a time of t seconds is:

vf = g * t

The above equation can be used to calculate the velocity of the object after any given amount of time when dropped from rest.

The distance that a free-falling object has fallen from a position of rest is also dependent upon the time of fall. This distance can be computed by use of a formula:

d = 0.5 * g * t2



The Big Misconception

Topic Sentence: During free-fall, all objects, regardless of mass, accelerate at the same rate (g-9.8 m/s/s).

g (9.8) is the same for all free-falling objects regardless of how long they have been falling, or whether they were initially dropped from rest or thrown up into the air.

Free-fall is the motion of objects that move under the sole influence of gravity; free-falling objects do not encounter air resistance, so all have acceleration of -9.8m/s/s. More massive objects will only fall faster if there is an appreciable amount of air resistance present.

The actual explanation of why all objects accelerate at the same rate involves the concepts of force and mass. The acceleration of an object is directly proportional to force and inversely proportional to mass. Increasing force tends to increase acceleration while increasing mass tends to decrease acceleration. The greater force on more massive objects is offset by the inverse influence of greater mass. Subsequently, all objects free fall at the same rate of acceleration, regardless of their mass.

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Lab: Freely Falling Object

Lauren Kostman
Partner: Jonathan Itskovich
October 4, 2011

Purpose: What is the acceleration of a falling body (due to gravity)?

Materials: ticker tape timer, ticker tape, masking tape, mass, clamp, meterstick

Hypotheses:
1.) The acceleration due to gravity is 9.8 m/s/s.
2.) The velocity-time graph of a free-falling object will have a slope of 9.8. The y-intercept is at zero (m).
3.) You can find the acceleration from this graph as acceleration is the slope of the v-t graph. This graph has a slope of 9.8, meaning an acceleration of 9.8 m/s/s.

Materials: ticker tape with stop timer, an object (mass=200g), meterstick/measuring tape, (masking) tape

*Mass of Object= 200 g

Data:
Position-Time Data Table:

x-t Graph:

Data Table:

v-t Graph:

Period 4 Class Results:

Sample Calculations:

Discussion Questions:

1. Does the shape of your v-t graph agree with the expected graph? Why or why not?

Yes, the shape of my v-t graph agrees with the expected graph because it is a straight line (has a constant slope), which represents a constant acceleration, and it moves in the positive direction. This means the object was moving away. The slope of the graph is 864, which represents the acceleration of the object (in cm/s/s).

1. Does the shape of your x-t graph agree with the expected graph? Why or why not?

Yes, the shape of my x-t graph agrees with the expected graph because it's a curved line, which shows it has an increasing, not constant, speed, which increases as it progresses in the positive direction, proving that it is moving away. The parabolic shape of the graph proves that the object was in fact accelerating, for a straight line on a position-time graph would represent constant speed, which in this case would've been incorrect for the situation.

1. How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)

Compared to the results of my classmates, my results were very similar to those of others. The percent difference between my results and those of my class is about 3.59%, which is a very small difference overall. Thus, this suggests that we had good results, in addition to the pretty low percent of error in our results, as well (11.9%).

1. Did the object accelerate uniformly? How do you know?

The object did accelerate uniformly, and this is known because the v-t graph is a straight line, and a straight line on such a graph represents an acceleration that is uniform. From equal amounts of time, the object speeds up in equal amounts. The percent error of our experiment was 11.9 percent, signifying that the mass did tend to accelerate at a constant, uniform rate.

1. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?

Realistically, acceleration due to gravity should not be greater than 9.8 m/s/s, unless there is another factor involved (that's pulling the object down at a faster rate). However, mistakes in the lab, such as crinkled ticker tape, or an issue with the ticker-tape stop timer can cause an error in the results, which could lead to an acceleration value being higher than it should be. If the value is lower than it should be, this could be due to a problem that occurred while the ticker tape was passing through the timer; if it got caught inside, it's movement would have been slowed down, thus causing a lower rate of acceleration during free-fall.

Analysis:
The shape of the position versus time graph is a parabolic line, which symbolizes that the mass had an increasing speed as it moved in the position direction, away from the origin. The trend line equation is y=433.47x^2+53.832x (form: y= Ax^2+Bx); much information can be taken from this equation. Half of the acceleration is equal to the "A" value, 433.47. The "B" value, which is 53.832, is equal to the initial velocity. Another way of viewing this equal (in terms of physics) is delta d= .5at^2 + vit. The r^2 value of this graph is .9999; this value represents how close (up to 100%) the points on the line fit the line of best fit (trend line). Evidently, our results were extremely close, given that this value was so close to 100%.
The shape of the velocity versus time graph is a straight line, which has a positive slope that increases in the positive direction, away from the origin. The equation for this graph is y=864x+55.07; the r^2 value is .99546. The slope of this line is 864, and since the slope on a v-t graph equals acceleration, that means that the acceleration of the mass was equal to 864 cm/s/s. The y-intercept of the graph is 55.07; this value was not set to zero because the initial velocity wasn't necessarily zero, because the instant the mass was dropped wasn't 100 percent accurate, although ideally it would be. Therefore, we cannot automatically set the y-intercept to zero. The r^2 value of our graph is representative of how close our graph fits to the trend line.

Conclusion:
Our first hypothesis was nearly correct; it was stated that acceleration due to gravity is 9.8 m/s/s (or 081 cm/s/s). Our results, however, showed acceleration due to gravity as being 864 cm/s/s, but due to our 11.9 percent error, this proves that our results were not completely accurate, and thus the 981 cm/s/s value for acceleration of gravity still holds it's truth. Secondly, it was believed that the slope of the line would be 9.8, with the y-intercept being at zero. As aforementioned, our acceleration was 864, rather than 981, thus this hypothesis wasn't completely accurate, as well. Furthermore, the y-intercept was not placed at zero, for our initial velocity could have been a tenth of a second before or after the dropping. Lastly, it was believed that the value for acceleration is represented by the slope on a velocity versus time graph, which is correct. Overall, the results derived from this lab where pretty accurate; we had a 11.9 % error, and a 3.59% difference from our period 4 class. The biggest issue seemed to be having the ticker tape get stuck in the timer; if this lab were to be redone, each person must be cautious of this prior to and during the dropping. Due to the friction caused by this mistake, the object wasn't actually in free-fall. Thus, in the future, people should try to hold the ticker tape carefully in a vertical position so it can slide quickly through the timer.

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